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1

One way of seeing it is by using the initial state and the final state as important markers of what went on. I'm going to assume the wall is a way to talk about an object that remains fixed in position (perhaps because it has a huge mass compared to the objects you are throwing at it, for example). Force is the rate of change in the quantity of motion over ...

-2

The Force on the wall is given by change in Momentum of the ball, Thus Force = Change in momentum F = mv Thus greater the speed, large the force will be.

2

A more mathematical way to think of it is that there are two relevant vectors: the initial velocity $\vec v$ and the initial acceleration $\vec a$. These span a two dimensional subspace (regardless of the number of dimensions in the full space). The solution will stay within this subspace. There is the degenerate case when $\vec v$ and $\vec a$ are not ...

3

As answered elsewhere, yes, the motion is confined to a plane. For me, analogy to cross products is less intuitive in higher dimensions. What makes it planar? Force dictates the second derivative. You always have a second order differential equation. All solutions to "nice" second order differential equations are defined by two functions. Below is a kinda-...

4

Tesseract's answer is correct. More conceptually, in general dimensions the angular momentum is what is known as a "two-form", which you can think of a plane with a magnitude and orientation of in-plane rotation. This works in any dimension, and in any dimension the conservation of this plane means that motion under a central force is confined to that plane. ...

10

In general, the angular momentum is defined as $\mathbf{L}=\mathbf{r}\wedge\mathbf{p}$. In our problem, switching to index notation, we have, in the CM frame, $$L_{\mu\nu}=r_\mu p_\nu-r_\nu p_\mu=\mu(r_\mu\dot{r}_\nu-r_\nu\dot{r}_\mu)$$ Now, since there is no external torque, angular momentum is constant. If we define a Cartesian coordinate system $(w,x,y,z)... 0 It holds true only along the contact normal direction. If$\boldsymbol{n}$is a vector normal to the contacting surfaces, then the law of contact states $$\boldsymbol{n} \cdot ( \boldsymbol{v}_{2f} - \boldsymbol{v}_{1f} ) = - \boldsymbol{n} \cdot ( \boldsymbol{v}_{2i} -\boldsymbol{v}_{1i} )$$ where$\cdot$is the vector dot product. The above projects ... 0 You are free to choose a reference frame, so you can choose the reference frame where one body is at rest. You now see the other body approaching you at a certain speed. The collision is elastic, so to conserve kinetic energy, after the collision you will indeed see the speed of recession equal to the speed of approach. 0 No one is shying away from Gravitational nonlocality. In the modern effective QFT language, all symmetry compliant terms should be included in the gravitational Lagrangian, which means we could have high-order derivative Lagrangian terms. On the other hand, given the negative mass dimension of gravitational constant, the renormalization enterprise would ... 1 OP's current$J^{\mu}$only satisfies the continuity equation on-shell. In contrast, the current from Noether's 2nd theorem is instead${\cal J}^{\mu}=\partial_{\nu}F^{\nu\mu}$, which satisfies the continuity equation off-shell, cf. e.g. my Phys.SE answer here. OP is correct: A gauge-symmetry [of the form$\alpha(x)=\varepsilon f(x)$, where$\varepsilon$is ... 1 The generators of the Lie algebra of a Lie group are defined (slight conventional dependence) as follows: If an element of the group$g( \boldsymbol \alpha) = \exp(i \boldsymbol \alpha \cdot \mathbf T)$is parameterised by a set of parameters$\alpha_i$then generator$T^i$can be found by differentiation: $$T^i = - i \frac{\partial}{\partial \alpha_i} g( \... 1 TL;DR: The first equation was incorrect and the example collision is problematic. You should generally use conservation of momentum. The first equation should be:$$v'_2 = \frac{2m_1}{m_1+m_2}v_1+\frac{m_2-m_1}{m_1+m_2}v_2$$Source However, this equation is only true for perfectly elastic collisions (i.e. total kinetic energy is conserved). Your example ... 1 Although the static friction does no work it does apply a force to the bottom of the rod (which accelerates the center of mass horizontally), and a torque. The normal force also produces a torque relative to te center of mass. Both of these torques change as the rod falls, and cause changes in the angular momentum. (The angular acceleration is a variable, ... 1 Energy conservation is a consequence of a system’s invariance under time translations. Momentum conservation is a consequence of a system’s invariance under spatial translations. Angular momentum conservation is a consequence of of system’s invariance under rotations. These relationships are all examples of Noether’s Theorem, named after the celebrated ... -1 The law of conservation of momentum states if there is no external forces act on system , the linear momentum remains always constant . So when a body is thrown upward neglecting external resistance, the linear momentum remains constant. The momentum before ball is thrown is zero so when the ball reaches upward there is gravity which is again external force ... 2 Yes, this is at present a low-quality derivation on Wikipedia. In particular the assembly m_0 - t \Delta m is not dimensionally consistent if t has any given units. One would instead write, say,$$\Delta v = \int_{t_0}^{t_1}\mathrm dt ~\frac{T}{m_0 - (m_0 - m_1)(t - t_0)/(t_1 - t_0)}.\tag{1}$$And then this expression is properly T/m_0 at t = t_0 and ... 0 Momentum is always conserved. If someone throws a ball upwards they impart a force on the ball. There is a corresponding reaction on the Earth through their feet. You can see this if you throw a ball upwards while standing on some scales- your weight will seem to momentarily increase. The momentum that the ball gains is offset by a change in the momentum of ... 0 Let the potential energy if the system be zero at infinity then gravitational potential energy of the system is given as$$E=\frac{-GMm}{r}$$You have missed the negative sign in your answer which accounts for the error.When the system is released from rest when it is at a separation of 2r then kinetic energy of the system increases which comes at the cost ... 3 You're missing a negative sign. The gravitational potential energy U of two masses m and M separated by a distance r is given by$$U=-G\frac{mM}{r}$$As the objects get closer together, the gravitational potential energy becomes more negative, which is another way to say that it decreases. As the potential energy decreases, the kinetic energy ... 1 You have a sign error. The potential energy for gravitation is lower the closer two objects are to each other, not higher. The potential energy of two objects separated by r is$$U = U_0 -\frac{GM_1 M_2}{r}$$where U_0 is the potential energy at infinite separation. U_0 is conventionally taken to be zero, but it doesn't really matter what value it ... 4 I am a bit confused because if that's the case, the atom will gain kinetic energy. In the way you've set it up (atom is initially at rest in our reference frame), then that is correct. Doesn’t that violate the conservation of energy? That depends on the energy inputs and outputs. We haven't described them completely yet. Since the energy is already ... 2 When the electron returns back to it’s ground state, a photon is emitted in a random direction Not exactly. If it's a spontaneous emission - then YES. If it's a stimulated emission - then NO, photon is emitted in the same direction of incident photon which has hit the atom. LASER or MASER works according to this principle. Btw, stimulated emission was ... 7 When a photon with the appropriate energy hits an atom, the electron will make a transition from the ground state to a excited state. This will make the potential energy of the atom higher. This is mostly correct, but it's not the potential energy of the atom which is higher, it's its internal energy Also, momentum is conserved, this is correct, but ... 2 Its because they are at rest so \gamma=1. The idea behind calculating threshold energy is to assume that all products are produced at rest in the COM frame. This way you get the energy just enough to produce them. Consider a process in lab frame$$A+B \rightarrow \sum_{i=1}^{n}C_i $$To calculate the threshold energy, we do the following: Assume the ... 0 If you are doing this in a simulation framework, instead of working with components on a rotated coordinate system and the problems that arise from measuring angles and having to decide which sign to use where, I suggest you work with vectors all of which need to be expressed on a common coordinate system. So here is the typical algorithm for handling ... 1 Here is the physics: Velocities along the "normal to plane of contact" (O_1O_2) act exactly like they do in 1D collisions. Velocities perpendicular to it stay unchanged Since you have rotated the velocities by \alpha, after rotation leave the y velocity components unchanged. After getting the post collision x velocities, add these vectorially to the ... 0 I suspect you have forgotten that \sigma_1 and \sigma_2 are both functions of x. So, following your notation :$$q(0) = \sigma_1(0)d^2=\frac{q}{d^2} d^2=qq(d)=\sigma_2(d)d^2=\frac{\kappa q}{d(d+(\kappa-1)d)}d^2=q$$And likewise it holds for any$x$. This makes sense because we explicitly used the following equation to derive$\sigma_1$and$\...

1

Just looking at steps 1 and 2, this doesn't look quite right to me. As I understand your diagram, $\alpha$ is the angle between the x-axis and a line that passes through the center of each circle. After the collision, you rotate the velocity of each object by the angle $\alpha$. In this model of mechanics, if one ball was directly above the other (so that \$\...

0

Since the total charge is constant I was expecting q(0)=q(d) This can't be. For the plates to remain at equipotential, some of the charge has to move from the part of the plate without inserted dielectric to the part of the plate with dielectric inserted in between. The total charge remains constant as you move the dielectric body. But the charge re-...

0

I think the intuitive reason for a 2d case is that you can resolve the velocity of the collision into two components, one being the head-on component and the other being parallel with the collision surface. The parallel component is going to be unchanged, and the head-on component will satisfy Newton's experimental law for a 1d collision, so the net effect ...

0

Normally the friction will not do any work during a quick collision of two hard bodies, like billiard balls. Nevertheless, it will have observable consequences in the form of rotation: part of the initial kinetic energy will be converted into rotational kinetic energy. I think if you assume perfect conservation of energy you might even be able to calculate ...

0

With friction operating in the system between the two bodies while impact(elastic),mechanical energy conservation law still holds good because in elastic impact friction doesn't do work as it is instantaneous(somewhat impulsive). Linear momentum may be conserved if there is no external force on two bodies but as friction(between colliding bodies) will be an ...

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